Answer :
Solution :
Given :
P (never married) = P = 0.44
Sample size, n = 15
Let [tex]$X \sim \text{Binomial}$[/tex] (n = 15, P = 0.44)
P(X = x) = [tex]$^nC_xP^x(1-P)^{n-x}; x=0, 1,....n$[/tex]
a). The probability that exactly two of them have been married is
P(X=2) = [tex]$^{15}C_2(0.44)^2(1-0.44)^{15-2}$[/tex]
[tex]$=\frac{15!}{2!(15-2)!}(0.44)^2(0.56)^{13}$[/tex]
[tex]$=\frac{15\times14}{2}\times (0.44)^2(0.56)^{13}$[/tex]
= 0.0108
b). That at most two of them have never been married.
[tex]$P(X \leq 2)= P(X=0)+P(X+1)+P(X=2)$[/tex]
[tex]$=^{15}C_0(0.44)^0(1-044)^{15-0}+^{15}C_1(0.44)^1(1-0.44)^{15-1}+^{15}C_2(0.44)^2(1-0.44)^{15-2}$[/tex]
[tex]$=(1)(1)(0.56)^{15}+(15)(0.44)(0.56)^{14}+(105)(0.44)^2(0.56)^{13}$[/tex]
= 0.000167+0.001969+0.010828
= 0.012964
= 0.0130
c). That at least 13 of them have been married.
P(married) = 1 - P(never married)
= 1 - 0.44
= 0.56
[tex]$P(X \geq 13)= P(X=13)+P(X=14)+P(X+15)$[/tex]
[tex]$=^{15}C_{13}(0.56)^{13}(1-0.56)^{15-13}+^{15}C_{14}(0.56)^{14}(1-0.56)^{15-14}+^{15}C_{15}(0.56)^{15}(1-0.56)^{15-15}$[/tex][tex]$=(105)(0.56)^{13}(0.44)^{2}+(15)(0.56)^{14}(0.44)+(1)(0.56)^{15}(1)$[/tex]
[tex]$=0.010828+0.001969+0.000167$[/tex]
= 0.012964
=0.0130